# Reconstructing a signal using an equilizer filter function

1. Mar 7, 2013

### Evo8

Reconstructing a signal using an equilizer "filter" function

1. The problem statement, all variables and given/known data

A low pass signal g(t) is sampled at a rate of $f_{s} > 2B$ needs to be reconstructed. Sampling interval is $T_s=\frac{1}{f_s}$

The reconstruction pulse to be used is

$$p(t)=∏(\frac{t}{T_s}-\frac{1}{2})$$

Specify the equalizer filter $E(f)$ to recover $g(t)$

2. Relevant equations
$$E(f)P(f)=0 / |f|>f_s-B$$
$$E(f)P(f)=T_s / |f|<B$$

$$E(f)=T_s \frac{ \pi f}{sin( \pi f T_p)} ≈ \frac{T_s}{T_p} \ \ |f|≤B$$

3. The attempt at a solution
There is one example in my text and it is not very clear at all.

I understand my "sampling interval" is $T_s=\frac{1}{f_s}$ What is $T_p$?

if my sampling frequency is greater then 2B then my last equation posted above does not apply right?

Im confused as where to start..

Thanks in advance for any help

Last edited: Mar 7, 2013
2. Mar 7, 2013

Change $$to $for your reconstruction pulse? And is that the world's largest pi in front of it or does it mean "product"? 3. Mar 7, 2013 ### Evo8 Sorry about the code error i guess i was in a hurry. Trying to balance work and homework. to answer your second question its both i guess? I never really have seen that notation before but now that you mention it, it makes more sense. The text im using and my professor for that matter both use the capital pi and the lower case pi but not interchangeably. 4. Mar 8, 2013 ### rude man Generally, upper-case pi denotes product and lower-case = 3.14.... Need to find out which was intended because this looks like a more challenging problem than the last one, even though it seems to use a similar sampling pulse train. 5. Mar 12, 2013 ### Evo8 Ok so i finally found what the books notation means when using$∏$Taken from the book: So I dont believe that it means the "product." that being said im not sure I fully see how to apply this.. Sorry for the late response. 6. Mar 12, 2013 ### rude man OK, I need to mull this one over. Like I said, more challenging than your last one. However, it seems that the reconstruction pulse is identical to the one used in your last problem [sampled g(t)], except due to the "1/2" in the PI function it looks like it's not centered at t = 0 but starts at t = 0. Stay tuned ... 7. Mar 12, 2013 ### berkeman ### Staff: Mentor Due to the advanced nature of thie problem, I'm going to grant rude_man some extra lattitude in providing help in this thread. 8. Mar 12, 2013 ### rude man In this case what is happening is that the signal is sampled every Ts seconds via what's called a zero-order hold process. You get a sequence of FLAT pulses, each representing the signal at the sampling instants. This is a pretty involved subject and even if I gave you hints you'd be hard-pressed to follow them I think. So instead I will direct you to a link that discusses the theory. Read the section starting with "Sampling with a zero-order hold" down to the "Interpolation" section. What you need to understand in this writeup is: 1. The meaning of your reconstruction pulse definition is that x(t) is zero-order-hold sampled. The ZOH output is xs(t). 2. The ZOH process is equivalent to delta-function sampling process followed by a filter with impulse response h0(t). The output of the h0(t) filter, which is the same as the output of the ZOH process, is inputed to a second filter hY(t). 3.hY(t) must be such that h0(t) * hY(t) = the impulse response of an ideal low-pass filter, the frequency spectrum of which H(jw) = 1 from w = 0 to w= 2πB and = 0 for w > 2πB. This fact is the result of Dirac-delta sampling and I cannot go into the details of this - it's elementary but too lengthy. (The beginning of this link does that to some extent). The frequency-domain of this ideal low-pass filter is H(jω). 4. So the filter transfer function HY(jw) you are looking for is their equation 2. The link is http://faraday.ee.emu.edu.tr/eeng420...s/lecture4.doc [Broken] Last edited by a moderator: May 6, 2017 9. Mar 12, 2013 ### rude man I tried the link & it didn't come back. If you send me you privatre e-mail I can e-mail you the file. If that's OK with Mr. Berkeman. Last edited: Mar 12, 2013 10. Mar 12, 2013 ### Evo8 I was able to download the lecture4.doc and pdf. http://faraday.ee.emu.edu.tr/eeng420/ If you click on hte URL above and then navigate to lecture notes on the left hand side then scroll down to lecture notes and select lecture4.doc or .pdf i beleive that is what you meant to link. I took another look through the book and did some more research online. I dont think this problem is meant to be so difficult.. According to my text (see screen shot below)$E(f)=T_s/P(f)$and$P(f)$is the fourier transform of my$p(t)$and the i found the fourier transform of a rectangular function is the following ∏(\frac{t}{\tau})= \tau sinc( \pi f \tau) and If\ \ g(t)\Leftrightarrow G(f) Then \ \ g(t-t_0) \Leftrightarrow G(f) \exp^{-j2 \pi ft_0} So I get$P(f)=T_{s} sinc( \pi f T_{s}) \exp^{-j \pi f}$And if$E(f)=\frac{T_s}{P(f)} \ \ and\ T_s=\frac{1}{f_s}\ \ where\ f_s>2B$Then I would get$E(f)=P(f)f_s$which yields E(f)=\frac{1}{2B} \ sinc( \pi f T_s) \exp^{-j \pi f} Maybe?..... However I feel that i handling the$f_s>2B$incorrectly.. 11. Mar 13, 2013 ### rude man EDIT: sorry, I didn't realize that the textbook p(t) had Tp in it but your problem has Ts instead. So just replace Tp with Ts since for the problem, Tp = Ts which makes p(t) = ∏(t/Ts - 1/2) So your text has already defined E(f) in terms of P(f) with Tp = Ts and all you need to do is perform E(f) = Ts/P(f), |f| <= B = arbitrary, B < |f| < 1/Ts - B = 0, |f| > 1/Ts - B. So, you already have your P(f) which is the result of the zero-order hold process I discussed earlier, so this is just math. Since this filter is NOT REALIZABLE it's hard to know to what extent you're supposed to simplify the expression for E(f). I mean, in a way the answer is there already. In real life no filter can have zero gain for |f| > 1/Ts - B or the ideal gain for |f| < B. All such E(f) are realizable approximations. The farther the spread between 2B and 1/Ts, the sloppier the filter can be, so in real life people sample at rates significantly greater than 1/Ts = 2B. Example: a CD player samples at a rate of at least 44 KHz even though the realistic cutoff frequency is about 15 KHz so the Nyquist rate would be 30 KHz. Now this you need to look at more carefully! (It would be nice if you could compute E(f) so easily! Hint: if y = a/x then y ≠ ax Since you were able to open the faraday file link I still think perusing it would be beneficial. You are going to have to deal with inverting P(f) one way or another. Last edited: Mar 13, 2013 12. Mar 14, 2013 ### Evo8 Yeah my E(f) is wrong right there. Just incorrect algebra. Not sure what I was going. Giving it another go I would get something like this$E(f)=\frac{T_s}{P(f)}=\frac{1}{P(f)f_s}## If this is correct then my filter transfer function should look something like this$$E(f)= \frac{1}{T_s \ sinc( \pi fT_s) \ e^{-j \pi f} (2B)}=\frac{1}{sinc(\frac{ \pi f}{2B} )e^{-j \pi f}}

And if I wanted to simplify further I would have to deal with inverting the P(f) as you mentioned. Or I would have attempted that before I "simplified" right?

13. Mar 14, 2013

### rude man

Not quite. Look again at the book's definition of E(f) and substitute Ts for Tp since their formula for E(f) was developed with arbitrary pulse width Tp whereas your pulse width = Ts = 1/fs.

You wind up with the 1/sinc()e-j function but you also need to specify the behavior of E(f) for the region f > B since your problem said that fs > 2B. Had it said fs = 2B there would not be a "flexible" region. The E(f) filter would have to pass everything for 0 < f < B and reject everything above f = B. If we sample faster than 2B we can allow "slop" in E(f) which is the "flexible" region. Then, for f > fs - B the filter E(f) needs to reject all signals since that is where the first "aliased" spectrum starts.

In the link I sent you their eq. 2 is identical with your E(f). Their H(jw) is the ideal low-pass filter which is H(jw) = 1, f < B and H(jw) = 0, f > B. That paper nicely shows how the aliased specrum develops as fs varies above and below 2B.

As I said, this E(f) is not realizable. Practical filters can never meet these E(f) specifications. So what is done in practice is to increase fs well above 2B in order to allow a wider "flexible" region and an acceptably low gain for when the first aliased spectrum intrudes, at f = fs - B.